If p, q, r are positive and are in AP, the roots of quadratic equation `px^(2) + qx + r = 0` are real for :

Options

A. `|(r)/(p) - 7| ge 4sqrt(3)`
B. `|(p)/(r) - 7| ge 4sqrt(3)`
C. all p and r
D. no p and r

Answer

To solve the problem, we need to determine the conditions under which the roots of the quadratic equation \( px^2 + qx + r = 0 \) are real, given that \( p, q, r \) are positive and in Arithmetic Progression (AP).

### Step-by-Step Solution:

1. **Understanding the condition for AP**:
Since \( p, q, r \) are in AP, we have:
\[ q = \frac{p + r}{2} \]

2. **Condition for real roots**:
The roots of the quadratic equation \( ax^2 + bx + c = 0 \) are real if the discriminant \( D \) is greater than or equal to zero. The discriminant \( D \) is given by:
\[ D = b^2 - 4ac \]
For our equation \( px^2 + qx + r = 0 \), this becomes:
\[ D = q^2 - 4pr \]

3. **Substituting \( q \)**:
Substitute \( q = \frac{p + r}{2} \) into the discriminant:
\[ D = \left(\frac{p + r}{2}\right)^2 - 4pr \]

4. **Expanding the discriminant**:
Expanding the expression, we have:
\[ D = \frac{(p + r)^2}{4} - 4pr \]
\[ D = \frac{p^2 + 2pr + r^2}{4} - 4pr \]
To combine the terms, we can express \( 4pr \) with a common denominator:
\[ D = \frac{p^2 + 2pr + r^2 - 16pr}{4} \]
\[ D = \frac{p^2 + r^2 - 14pr}{4} \]

5. **Setting the discriminant greater than or equal to zero**:
For the roots to be real, we need:
\[ p^2 + r^2 - 14pr \geq 0 \]

6. **Rearranging the inequality**:
Rearranging gives:
\[ p^2 - 14pr + r^2 \geq 0 \]

7. **Factoring the quadratic**:
We can factor this expression:
\[ (p - 7r)^2 \geq 0 \]
This inequality holds for all real numbers, and it equals zero when \( p = 7r \).

8. **Conclusion**:
Since \( p, q, r \) are positive, the roots of the quadratic equation \( px^2 + qx + r = 0 \) are real if:
\[ \frac{r}{p} \geq 7 \quad \text{or} \quad \frac{r}{p} \leq 7 \]